Let the vector space V
contain the vectors 5, negative 5, negative 3,
and 1, 4, negative 5.
Which of the following are vectors
in the space spanned by V?
In order for a vector to be in the space spanned by V,
it must be a linear combination of the vectors in V.
If the vector could not be written
as a linear combination of the vectors in V,
it is not in the space spanned by V.
So there's a couple approaches
we can take to solve this problem.
One approach would be to let C sub 1 and C sub 2
be scalars and set up an equation
using each of the given vectors from the list.
Where we have C sub 1 times the first vector in V
plus C sub 2 times the second vector in V
equals each vector from the list.
From there, we would write a system of equations
and then try to solve the system.
If the system has a solution,
then the vector is a linear combination of the vectors in V
and is in the space spanned by V.
If the system doesn't have a solution,
then the vector is not a linear combination
of the vectors in V,
and therefore, is not in the space spanned by V.
But for this problem, let's take a different approach.
Let's set up the vector equation,
where we let the right side be the generic vector
with components X sub 1, X sub 2, and X sub 3.
From here, breaking the system of equations,
we have 5C sub 1 plus C sub 2 equals X sub 1,
negative 5C sub 1 plus 4C sub 2 equals X sub 2.
And then for the last equation,
we have negative 3C sub 1 minus 5C sub 2
equals X sub 3.
From here, we'll write an augmented matrix
and then write the augmented matrix
in reduced row-echelon form.
What's important to recognize here
is that because we have three equations
with only two unknowns
in order for the system to have a solution
in reduced row-echelon form,
we must have a row of 0s in row three,
because if we don't then the vector X sub 1,
X sub 2, X sub 3 would not be a linear combination
of the given vectors in V,
and therefore, the vector would not be
in the space spanned by V.
Let's begin to write the augmented matrix
in reduced row-echelon form.
Let's get a 0 in row two column one.
I'm replacing row two with row one plus row two.
Let's also going to zero and row three column one.
We take a look at row one and row three,
the least common multiple of 5 and 3 is 15.
Let's replace row three with 3 times row one
plus 5 times row 3.
Row one stays the same.
Replacing row two with row one plus row two,
we have negative 5 plus 5 that's zero.
Next, we have 1 plus 4, which is 5.
And then we have X sub 1 plus X sub 2.
Now, we replaced row three with 3 times row one
plus 5 times row three.
3 times 5 plus 5 times negative 3 is 0.
3 times 1 plus 5 times negative 5 is negative 22.
And then finally, we have 3 times X sub 1
plus 5 times X sub 3, which is 3X sub 1 plus 5X sub 3.
And now let's get a 0 in row three column two.
Looking at row two in row three,
the least common multiple of 5 and 22 is 110.
Let's replace row three with 22 times row two,
and then plus 5 times row three.
Notice 22 times 5 is 110 plus 5 times negative 22
is negative 110 giving us a sum of 0.
Row one and row two stay the same.
And now we're replacing row three with 22 times row two
plus 5 times row three.
We have 22 times 0 plus 5 times 0, which is 0.
22 times 5 plus 5 times negative 22 is 0.
And then finally, we have 22 times the quantity X sub 1
plus X sub 2 plus 5 times the quantity 3X sub 1
plus 5X sub 3.
Let's work this out on this side.
We have 22 times the quantity X sub 1 plus X sub 2,
and then plus 5 times the quantity 3X sub 1
plus 5X sub 3.
Let's close the parentheses.
We have 22X sub 1 plus 22X sub 2 plus 15X sub 1
plus 25X sub 3.
And we do have two like terms here.
We have 2X sub 1 terms,
which gives us 37X sub 1 plus 22X sub 2
plus 25X sub 3.
This is the entry in row three column three.
Now we can go ahead and stop here.
Remember, in order for the system to have a solution,
because we have three equations with only two unknowns,
the third row must be a row of 0s,
which indicates that 37X sub 1 plus 22X sub 2
plus 25X sub 3 must equal 0.
Where X sub 1, X sub 2, and X sub 3
are the components of the vectors.
So now that we have this equation, we can simply substitute
in the components of each of the vectors from the list
and see which of them satisfy the equation.
If the vector doesn't satisfy the equation,
it is not a linear combination of the vectors in V
and also not in these spaces spanned by V.
If the vector does satisfy the equation,
it can be written as a linear combination
of the vectors in V,
and therefore, is in the space spanned by V.
To save some time I've already set most of this up
where here's the equation that must be satisfied
in order for the vector to be
a linear combination of the vectors in V,
and therefore, in the space spanned by V.
So I've listed the five vectors
and already performed the substitution.
However, we should recognize the first vector,
5, negative 5, negative 3
is already one of the vectors in V,
and therefore, it is a linear combination
of the two vectors.
Either way though, notice how it does satisfy the equation,
because it does equal 0.
Looking at the second vector of negative 1, negative 4, 5,
and performing substitution.
We also get 0, and therefore, these first two vectors
are linear combinations of the vectors in V,
and therefore, in the space spanned by V,
but don't have the three remaining vectors
do not satisfy the equation.
For the third vector, we get negative 4,
which doesn't equal 0.
For the fourth vector, we get negative 232,
which doesn't equal 0.
And for the fifth vector, we get negative 216,
which again, doesn't equal 0.
So because these last three vectors
do not satisfy the equation,
we know they can not be written
as linear combinations of the vectors in V,
and therefore, the vectors are not
in the space spanned by V.
So we only checked the first two vectors.
I hope you found this helpful.